Question

In the circuit of given figure, the final voltage drop across the capacitor $ C $ is

• A. $ \frac{Vr_{1}}{r_{1}+r_{2}} $
• B. $ \frac{Vr_{2}}{r_{1}+r_{2}} $
• C. $ \frac{V\left(r_{1}+r_{2}\right)}{r_{2}} $
• D. $ \frac{V\left(r_{1}+r_{2}\right)}{r_{1}+r_{2}+r_{3}} $

Answer 1

Answer: (B)

In steady state, the capacitor arm produce an infinite resistance. So, the potential difference across $C$ is that across $r_{2}$.
So, current through $r_{2}=\frac{V}{\left(r_{1}+r_{2}\right)}$
Potential difference across $r_{2}=\frac{V\left(r_{2}\right)}{\left(r_{1}+r_{2}\right)}$